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# JEE Main Study Notes for Statistics: Important Formulas, Short-Cut Methods & Previous Year Solved Questions Statistics is one of the topics that consistently appears in JEE Main paper every year. At least one or two questions are fetched from this section. Statistics normally means plotting or arranging facts, observations and information that come from an investigation in a meaningful manner. The information received (termed as data) can be in the form of individual series, discrete frequency, or continuous frequency. To understand such a data, we use measures of central tendency that reflects the behaviour of the whole data. Read JEE Main Mathematics Syllabus

• The probability of getting at least one question from Statistics is 100%. In the past 10 years JEE Main question paper, one question from statistics has appeared.
• Though the weightage of the section is around 3.33%, the recurrency of the section makes it a prominent topic for JEE Main.

Read the article to know about important formulas under the topic, basic concepts, short-cut methods, and solved questions from previous years’ JEE Main papers.

### Measures of Central Tendency

A Central Tendency is a central value of a probability distribution. The measures indicate where most values fall in the distribution. In statistics, there are two types of averages under which the central measures of tendency fall:

• Mathematical Averages
• Arithmetic mean or Mean
• Geometric mean
• Harmonic mean
• Averages of position
• Median
• Mode

### Arithmetic Mean

#### Arithmetic Mean for Unclassified Data

If n numbers be x1, x2, x3 ........ xn then their arithmetic mean is

#### Arithmetic Mean for Frequency Distribution

Let f1, f2, ........, fn be corresponding frequencies of x1, x2, .......xn. Then

#### Arithmetic Mean for Classified Data

Class Mark of the class-interval a – b,

For a classified data, we take the class-marks x1, x2, ...., xn of the classes as variables and

### Geometric Mean

If x1, x2, ......, xn be n values of the variable then

For Frequency Distribution, where

or

### Harmonic Mean

For Frequency Distribution, , where

### Short-cut Method to Calculate Arithmetic Mean

#### Short-cut Method for Simple Distribution

Where, a = assumed mean, d = x – a, n = no. of terms

#### Short-cut Method for Unclassified Frequency Distribution

Where a = assumed mean, d = x – a, f = frequency of variable x

#### Short-cut Method for Classified Frequency Distribution

Where a = assumed mean, d = x – a, x = class-mark of the class-interval,  f = frequency of the class interval

### Median

Median is the middle-most value of the data given, when arranged in ascending or descending order. If the question given is in the form of ungrouped data, we first arrange the data into ascending or descending order and then identify the Median by applying the formulas in accordance with the data.

#### Determination of Median for Simple Distribution

Arrange the data provided in ascending or descending order and then find the n (number of terms).

• If the number of terms are odd then term is the Median
• If number of terms are even then there are two middle terms namely,

and term. Hence, Median = Mean of and terms.

#### Determination of Median for Unclassified Frequency Distribution

• First we need to find , where
• Calculate the cumulative frequency of each value of the variable and then take value of the variable which is equal to or just greater than . The value of the chosen variable is the median.

#### Determination of Median for Classified Data (Class Limit & Boundary)

At times the questions are asked in the form of Overlapping intervals, e.g. 10 – 20, 20 – 30, 30 – 40, where the Upper limit for 10 - 20 interval = 20, and Lower limit = 10

For Non-overlapping intervals, e.g. 10 - 19, 20 - 29, Upper boundary for 10 - 19 = = lower boundary of 20 – 29.

#### Determination of Median by Graph

To determine median with the help of graphs, draw the “less than” ogive and “more than” ogive for the distribution. The abscissa of the point of intersection of these ogives is the median.

### Mode

Mode is the variable which occurs maximum number of times in a given series of elements. It is the value at the point in the distribution where items tend to be most heavily concentrated.

#### Mode for a Raw Data

Mode from the following numbers of a variable 70, 80, 90, 96, 70, 96, 96, 90 is 96 as 96 occurs maximum number of times.

#### Mode for Unclassified Frequency Distribution

In the above given data, the frequency that has appeared maximum number of times is 5, with 13 occurrences. Hence, the mode is 5.

#### Mode for Classified Distribution

Class that has maximum frequency is called the MODAL CLASS and the middle point of the modal class is called the CRUDE MODE. The class just before the modal class is called PRE-MODAL CLASS and the class after the modal class is called the POST-MODAL CLASS

Determination of mode for classified Data (continuous distribution)

• l = lower limit of the modal class
• f0 = frequency of the modal class
• f–1 = frequency of the pre-modal class
• f1 = frequency of the post-modal class
• i = length of the class-interval.

### Dispersion

Dispersion normally means the extent to which the numerical data tends to spread. It is also called as Variation. Its measurement is called Deviation. There are four Measures (methods) of Dispersion:

• Mean Deviation
• Standard Deviation
• Range
• Quartile Deviation

#### Mean Deviation

Also called as Mean Absolute Deviation, it is the mean of the absolute deviations of a set of data from its mean denoted by or or

• For simple (discrete) distribution

where n = no. of terms, z = A or M or M0

• For Unclassified frequency distribution
• For classified distribution

where x stands for class-mark

#### Standard Deviation

It measures the dispersion of the dataset relative to its mean. It is calculated as the sqaure root of the variance. (The square of S.D., i.e., σ2 is called the Variance.)

• For simple (discrete) distribution
• For frequency distribution
• For classified data

where x = class-mark of the interval.

### Solved Previous Year Questions on Statistics

Question: Compute the median from the following table.

Marks obtainedNo. of students
0-102
10-2018
20-3030
30-4045
40-5035
50-6020
60-706
70-803

Ans:

Marks obtainedNo. of studentsCumulative frequency
0-1022
10-201820
20-303050
30-404595
40-5035130
50-6020150
60-706156
70-803159

N = Σf = 159 (Odd number)

Median is [1 / 2] (n + 1) = [1 / 2] [(159 + 1)] = 80th value, which lies in the class [30 – 40] (see the row of cumulative frequency 95, which contains 80).

Hence the median class is [30 – 40].

We have l = Lower limit of median class = 30

f = frequency of median class = 45

C = Total of all frequencies preceding median class = 50

i = width of class interval of median class=10

Required median = l + ([N / 2 − C] / f) * i

= 30 + ([159 / 2 − 50] / 45) × 10

= 3 + 295 / 45

= 36.55

Question: The mean and S.D. of the marks of 200 candidates were found to be 40 and 15 respectively. Later, it was discovered that a score of 40 was wrongly read as 50. The correct mean and S.D. respectively are ______________.

Ans: Corrected Σx = 40 × 200 − 50 + 40 = 7990

Corrected x bar = 7990 / 200 = 39.95

Incorrect Σx2 = n [σ+ (\bar{x}^2xˉ2) = 200 [152 + 402] = 365000

Correct Σx= 365000 − 2500 + 1600 = 364100

Corrected σ = \sqrt{\frac{364100 }{ 200}}200364100​​ − (39.95)2

= √(1820.5 − 1596)

= √224.5

= 14.98

Question: The average of n numbers x1, x2,……. xis M. If xn is replaced by x′, then what is the new average?

Ans: M = \frac{x_1, x_2,……. x_n}{n}nx1​,x2​,…….xn​​ i.e.,

nM = x1, x2,……. xn−1 + xn

nM – x= x1, x2,……. xn−1

[nM – xn + x′] / n = [x1, x2,……. xn−1 + x′] / n

New average = [nM − xn + x′] / n

Question: The following data gives the distribution of the height of students.

 Height (in cm) Name of students 160 150 152 152 161 154 155 12 8 4 4 3 3 7

What is the median of the distribution?

Ans: Arranging the data in ascending order of magnitude, we obtain

 Height (in cm) Number of students Cumulative frequency 150 152 154 155 156 160 161 8 4 3 7 3 12 4 8 12 15 22 25 37 41

Here, the total number of items is 41 i.e., an odd number.

Hence, the median is [(41 + 1) / 2]th i.e., 21st item.

From the cumulative frequency table, we find that median i.e., 21st item is 155.

(All items from 16 to 22nd are equal, each 155).

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